✅ 1. CACHE MAPPING NUMERICAL (Direct Mapping) NUMERICALS

 

✅ 1. CACHE MAPPING NUMERICAL (Direct Mapping)

Q1.

A computer has:

• Main memory = 4K words

• Cache = 128 words

• Block size = 4 words

Find:

1. No. of blocks in main memory

2. No. of lines in cache

3. Mapping of memory block 205 to cache line.

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Solution

1) Blocks in main memory

Main memory size = 4K = 4096 words
Block size = 4 words

$$\text{No. of blocks} = \frac{4096}{4} = 1024$$

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2) Cache lines

Cache size = 128 words
Block size = 4 words

$$\text{Cache lines} = \frac{128}{4} = 32$$

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3) Mapping of block 205

Direct mapping:

$$\text{Cache line} = \text{Block number} \mod \text{No. of lines}$$ $$205 \mod 32 = 13$$

✔ Answer:

Block 205 maps to cache line 13.

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✅ 2. SET-ASSOCIATIVE CACHE NUMERICAL

Q2.

8 KB cache, block size = 64 bytes, 4-way set associative.
Find number of sets.

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Solution

Cache size = 8 KB = 8192 bytes
Block size = 64 bytes

1 set has 4 blocks (because 4-way).

Total blocks =

$$\frac{8192}{64} = 128$$

No. of sets =

$$\frac{128}{4} = 32$$

✔ Answer:

32 sets

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✅ 3. MEMORY INTERLEAVING

Q3.

A system has 4 memory banks using lower-order interleaving.
Words are stored starting from address 0.
Find the bank number for address: 37.

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Solution

Lower-order interleaving:

$$\text{Bank} = \text{Address} \mod \text{No. of banks}$$

Here:
37 mod 4 = 1

✔ Answer:

Address 37 is in Bank 1.

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✅ 4. EFFECTIVE ACCESS TIME (EAT)

Q4.

Hit ratio = 90%
Cache access time = 10 ns
Main memory access = 100 ns

Find EAT.

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Solution

$$EAT = (Hit\ Ratio \times Cache\ Time) + (Miss\ Ratio \times Memory\ Time)$$

Hit ratio = 0.9
Miss ratio = 1 – 0.9 = 0.1

$$EAT = (0.9 \times 10) + (0.1 \times 100)$$ $$= 9 + 10 = 19\ \text{ns}$$

✔ Answer:

19 ns

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✅ 5. PIPELINE THROUGHPUT

Q5.

A pipeline has 5 stages, each taking 20 ns.
Calculate time to execute 10 instructions.

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Solution

Pipeline time =

$$kT + (n - 1)T$$

k = stages = 5
T = 20 ns
n = instructions = 10

$$= 5(20) + 9(20)$$ $$= 100 + 180 = 280\ \text{ns}$$

✔ Answer:

280 ns

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✅ 6. BINARY → EXCESS-3 NUMERICAL

Q6.

Convert 75 (decimal) into Excess-3 code.

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Solution

Digits: 7 and 5

Step 1 → Convert to BCD:
7 → 0111
5 → 0101

Step 2 → Add 3 (0011) to each:

0111 + 0011 = 1010
0101 + 0011 = 1000

✔ Answer:

75 = 1010 1000 (Excess-3)

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✅ 7. BINARY MULTIPLICATION (Booth’s Algorithm)

Q7.

Multiply: (+7) × (−3) using Booth’s algorithm.
(Short exam-style solution.)

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Solution Summary

7 = 0111
−3 = 1101 (2’s complement)

Perform Booth steps → final answer:

✔ Answer:

(+7) × (−3) = −21
Binary result: 1010 1011

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✅ 8. ADDRESSING MODE NUMERICAL

Q8.

Effective address = ?
Instruction: MOV A, M
HL register pair = 2050H

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Solution

Indirect addressing:
Address = content of HL.

✔ Answer:

EA = 2050H

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✅ 9. MEMORY SIZE CALCULATION

Q9.

How many bytes in a memory with 16-bit address bus?

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Solution

No. of addressable locations =

$$2^{16} = 65536$$

Each location stores 1 byte.

✔ Answer:

64 KB

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✅ 10. CACHE TAG CALCULATION

Q10.

Find tag, set, and block offset for:

• Cache = 4 KB

• Block size = 32 bytes

• 2-way associative

• 16-bit address

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Solution Steps

1. Cache size = 4096 bytes
   Block size = 32 →
   Blocks = 4096 / 32 = 128 blocks

2. 2-way set associative →
   Sets = 128 / 2 = 64 sets

3. Set bits = log₂(64) = 6 bits

4. Offset bits = log₂(32) = 5 bits

5. Tag bits = 16 – (6 + 5) = 5 bits

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✔ Final Answer:

• Tag = 5 bits

• Set = 6 bits

• Offset = 5 bits